N2O5 dissolved in CCl4 decomposes to give N2O4 and O2. The reaction is first order with a half-life of 1234 seconds. How long will it take, in seconds, for the concentration of N2O5 to fall to 1.1% of its initial value?

the fraction left after t seconds is (1/2)^(t/1234)

so, solve

(1/2)^(t/1234) = 0.011
t = 8028.84

makes sense: .011 is around 1/64, or 6 half-lives

t=8028.84

8028.84

To find the time it takes for the concentration of N2O5 to fall to 1.1% of its initial value, we first need to determine the reaction rate constant (k) for the decomposition of N2O5.

Since the reaction is first order, the rate equation can be written as:

Rate = k[N2O5]

The half-life (t1/2) of a first-order reaction is given by the equation:

t1/2 = 0.693 / k

Given that the half-life (t1/2) is 1234 seconds, we can rearrange the equation to solve for the rate constant (k):

k = 0.693 / t1/2

Let's calculate the rate constant (k):

k = 0.693 / 1234
k ≈ 5.62 x 10^(-4) s^(-1)

Now, we can use the rate equation to find the time it takes for the concentration of N2O5 to fall to 1.1% of its initial value. The concentration at any given time (t) can be calculated using the equation:

[N2O5]t = [N2O5]0 * e^(-kt)

Where:
[N2O5]t is the concentration of N2O5 at time t,
[N2O5]0 is the initial concentration of N2O5,
e is the base of the natural logarithm (approximately 2.718),
k is the rate constant, and
t is the time.

In this case, we want to find the time (t) when [N2O5]t is equal to 0.011 times [N2O5]0:

[N2O5]t = 0.011 * [N2O5]0

Substituting the given values, we get:

[N2O5]0 * e^(-kt) = 0.011 * [N2O5]0

Now, we can cancel out [N2O5]0:

e^(-kt) = 0.011

Taking the natural logarithm (ln) of both sides:

-kt = ln(0.011)

Solving for t:

t = -ln(0.011) / k

Let's plug in the value of k:

t = -ln(0.011) / (5.62 x 10^(-4) s^(-1))

Calculating t:

t ≈ 2540 seconds

Therefore, it will take approximately 2540 seconds for the concentration of N2O5 to fall to 1.1% of its initial value.